ECE291 Computer Engineering II Lecture 6

ECE291Computer Engineering IILecture 6 Josh Potts University of Illinois at Urbana- Champaign

Outline • Program organization • NASM directives • Multiplication • Division • Macros

Program Organization • Create block structure and/or pseudocode on paper to get a clear concept of program control flow and data structures • Break the total program into logical procedures/macros • Use jumps, loops, etc. where appropriate • Use descriptive names for variables • noun_type for types • nouns for variables • verbs for procedures/functions

Debugging Hints • Good program organization helps • Programs do not work the first time • Strategy to find problems • Use DEBUG breakpoints to check program progress • Use COMMENT to temporarily remove sections of code • "print" statements announce milestones in program • Test values/cases • Try forcing registers/variables to test output of a procedure • Use "print" statements to display critical data • Double-check your own logic (Did you miss a special case?) • Try a different algorithm, if all else fails...

NASM Directives • Includes %include “drive:\path\filename” • Definitions • DB/RESB define/reserve byte (8bits) • DW/RESW define/reserve word (16 bits) • DD/RESD define/reserve double word (32 bits) • EQU names a constant • Labels • “.” prefixed “local” to previous non-dotted label • “:” suffix (optional, not required)

NASM Directives(cont.) • Macros • Instead of using procedures, which require both stack and time resources, macros are fast and flexible • Advantages: • speed; no call instruction • readability - easier to understand program function • Drawbacks - • space using the macro multiple times duplicates the code • tricky to debug (in particular when you have nested macros) • Procedures • “name” • use “.” (local) labels inside procedure (helps keep labels unique)

NASM Directives (cont.) • References to procedures • EXTERN “name” – give you access to procedures/variables in other files • GLOBAL “name” – makes your procedures/variables available to other files • Segment definition • SEGMENT “name” [STACK]

Example Program Structure ; ECE291:MPXXX ; In this MP you will develop program which take input ; from the keyboard………… ;====== Constants================================================= ;ASCII values for common characters CR EQU 13 LF EQU 10 ESCKEY EQU 27 ;====== Externals================================================= ; -- LIB291 Routines extern dspmsg, dspout, kbdin extern rsave, rrest, binasc

Example Program Structure (cont.) ;==== LIBMPXXX Routines (Your code will replace calls to these functions) extern LibKbdHandler extern LibMouseHandler extern LibDisplayResult extern MPXXXXIT ;====== Stack ===================================================== stkseg segment stack ; *** STACK SEGMENT *** resb 64*8 ; 64*8 = 512 Bytes of Stack stacktop: ;====== Begin Code/Data============================================ segment CODE ; *** CODE SEGMENT ***

Example Program Structure (cont.) ;====== Variables================================================= inputValid db 0 ; 0: InputBuffer is not ready ; 1: InputBuffer is ready ;-1: Esc key pressed operandsStr db 'Operands: ','$' OutputBuffer 16 times db 0 ; Contains formatted output db ‘$’ ; (Should be terminated with '$') MAXBUFLENGTH EQU 24 InputBuffer MAXBUFLENGTH times db 0 ; Contains one line of db ‘$’ ; user input graphData %include “graphData.dat” ; data GLOBAL OutputBuffer, inputValid, operandsStr GLOBAL graphData

Example Program Structure (cont.) ;====== Procedures =========================================== KbdHandler <Your code here> MouseHandler <Your code here> DisplayResult <Your code here> ;====== Program Initialization =============================== ..start: mov ax, cs ; Use common code & data segment mov ds, ax mov ax, stkseg ; Initialize stack mov ss, ax mov sp, stacktop

Example Program Structure (cont.) ;====== Main Procedure ======================================== MAIN: MOV AX, 0B800h ;Use extra segment to access video screen MOV ES, AX <here comes your main procedure> CALL MPXXXXIT ; Exit to DOS

Multiplication • The product after a multiplication is always a double-width product, e.g, • if we multiply two 16-bit numbers , they generate a 32-bit product • unsigned: (216 - 1) * (216 - 1) = (232 - 2 * 216 + 1 < (232 - 1) • signed: (-215) * (-215) = 230 < (231 - 1) • overflow cannot occur • Modification of Flags • Most flags are undefined after multiplication • O and C flags clear to 0 if the result fit into half-size register • e.g., if the most significant 16 bits of the product are 0, both flags C and O clear to 0

Multiplication (cont.) • Two different instructions for multiplication • MULMultiply unsigned • IMUL Integer Multiply (2’s complement) • Multiplication is performed on bytes, words, or double words • Which operation to perform depends on the size of the multiplier • The multiplier can be any register or any memory location mul cx ; AX * CX (unsigned result in DX--AX);imul word [si] ; AX * [word content of memory location ; addressed by SI] (signed productin DX--AX)

Multiplication(16 bit) The use of the AX (and DX) registers is implied!!!!! Multiplicand AX Multiplier (16-bit register, 16-bit memory variable) DX, AX = PRODUCT (High word in DX : Low word in AX)

Multiplication (cont.) • 8086/8088 microprocessors do not allowto perform immediate multiplication • 80286, 80386, and 80486 allow the immediate multiplication by using a special version of the multiply instruction • Immediate multiplication must be signed multiplication and contains three operands • 16-bit destination register • register or memory location that contains 16-bit multiplicand • 8-bit or 16-bit immediate data used as a multiplier imul cx, dx, 12h ;multiplies 12h * DX and leaves ;16-bit signed product in CX

Multiplication • 8-bit multiplication Multiplicand AL Multiplier (8-bit register, 8-bit memory variable) AX PRODUCT • 32-bit multiplication Multiplicand EAX Multiplier (32-bit register, 32-bit memory variable) EDX, EAX PRODUCT (High word in EDX : Low word in EAX) • 32-bit multiplication is available only on 80386 and above

Binary Multiplication • Long Multiplication is done through shifts and additions • This works if both numbers are positive • To multiply a negative numbers, the CPU will store the sign bits of the numbers, make both numbers positive, compute the result, then negate the result if necessary 0 1 1 0 0 0 1 0 (98) x 0 0 1 0 0 1 0 1 (37) ------------------------- 0 1 1 0 0 0 1 0 0 1 1 0 0 0 1 0 - - 0 1 1 0 0 0 1 0 - - - - - (3626)

Division • X / Y = Q; R X Dividend Y Divisor Q Quotient R Remainder Note: Remainder has the same sign as X (Dividend) Examples (Signed Integers) X / Y Q R 9 / 4 2 1 -9 / 4 -2 -1 9 / -4 -2 1 -9 / -4 2 -1

Division (cont.) • Two different instructions for division • DIVDivision unsigned • IDIV Integer Division (2’s complement) • Division is performed on bytes, words, or double words • Which operation to perform depends on the size of the divisor • The dividend is always a double-width dividend that is divided by the operand (divisor) • The divisor can be any register or any memory location

Division(32-bit/16-bit) The use of the AX (and DX) registers is implied!!!!! Dividend DX, AX (high word in DX, low word in AX) Divisor (16-bit register, 16-bit memory variable) Quotient AX Remainder DX

Division (cont.) • 16-bit/8-bit Dividend AX Divisor (8-bit register, 8-bit memory variable) Quotient AL Remainder AH • 64-bit/32-bit Dividend EDX, EAX (high double word in EDX, low double word in EAX) Divisor (32-bit register, 32-bit memory variable) Quotient EAX Remainder EDX • Available on 80386 and above

Division (cont.) • Division of two equally sized words • Prepare the dividend • Unsigned numbers: move zero into high order-word • Signed numbers: use signed extension (implicitly uses AL, AX, DX registers) to fill high-word with ones or zeros • CBW(convert byte to word) AX = xxxx xxxx snnn nnnn (before) AX = ssss ssss snnn nnnn (after) • CWD(convert word to double) DX:AX = xxxx xxxx xxxx xxxx snnn nnnn nnnn nnnn (before) DX:AX = ssss ssss ssss ssss snnn nnnn nnnn nnnn (after) • CWDE(convert double to double-word extended) - 80386 and above

Division (cont.) • Flag settings • none of the flag bits change predictably for a division • A division can result in two types of errors • divide by zero • divide overflow (a small number divides into a large number), e.g., 3000 / 2 • AX = 3000; • Devisor is 2 => 8 bit division is performed • Quotient will be written to AL => but 1500 does not fit into AL • consequently we have divide overflow • in both cases microprocessor generates interrupt (interrupts are covered later in this course)

Division (Example) Division of the byte contents of memory NUMB by the contents of NUMB1 Unsigned MOV AL, [NUMB] ;get NUMB MOV AH, 0 ;zero extend DIV byte [NUMB1] MOV [ANSQ], AL ;save quotient MOV [ANSR], AH ;save remainder Signed MOV AL, [NUMB] ;get NUMB CBW ;signed-extend IDIV byte [NUMB1] MOV [ANSQ], AL ;save quotient MOV [ANSR], AH ;save remainder

Division (cont.) • What do we do with remainder after division? • use the remainder to round the result • drop the remainder to truncate the result • if the division is unsigned, rounding requires that remainder is compared with half the divisor to decide whether to round up the quotient • e.g., sequence of instructions that divide AX by BL and round the result DIV BL ADD AH, AH ;double remainder CMP AH, BL ;test for rounding JB .NEXT INC AL .NEXT:

ECE291 Computer Engineering II Lecture 6